2014-11-23 81 views
1

我使用GraphViz库,并且在某些情况下,它回退了一个以元组为键的字典。使用元组作为字典键

{(4, 7): 0, (2, 6): 1, (10, 4): 1, (5, 11): 1, (4, 5): 1, 
(2, 8): 0, (8, 11): 0, (10, 0): 1, (6, 11): 1, (9, 11): 1, 
(1, 9): 0, (10, 1): 0, (7, 11): 1, (0, 9): 1, (3, 7): 1, 
(10, 3): 1, (10, 2): 1} 

出于某种原因,我想获得在元组第二个数字: 第一号== 10 值== 1

我试着用(10,)访问字典,但我认为这种语法是不允许在Python中。

答案应该是:[4 ,0 ,3 , 2]

+2

'1'不应该是结果的一部分,应该吗? – 2014-11-23 10:48:02

+0

是的,谢谢:) – 2014-11-23 10:51:03

回答

6

你必须遍历字典,例如:

In [1]: d = {(4, 7): 0, (2, 6): 1, (10, 4): 1, (5, 11): 1, (4, 5): 1, 
    ...: (2, 8): 0, (8, 11): 0, (10, 0): 1, (6, 11): 1, (9, 11): 1, 
    ...: (1, 9): 0, (10, 1): 0, (7, 11): 1, (0, 9): 1, (3, 7): 1, 
    ...: (10, 3): 1, (10, 2): 1} 

In [2]: [b for (a, b), v in d.items() if a == 10 and v == 1] 
Out[2]: [4, 0, 3, 2] 
1
result=[] 
for key in your_dict.keys(): 
    if key[0]==10 and your_dict[key]==1: 
    result.append(key[1]) 
1

(10,)是一个完全有效的语法,但是这将提高KeyError这里。要获得所需的输出,您必须在此使用循环:

>>> d = {(4, 7): 0, (2, 6): 1, (10, 4): 1, (5, 11): 1, (4, 5): 1, 
... (2, 8): 0, (8, 11): 0, (10, 0): 1, (6, 11): 1, (9, 11): 1, 
... (1, 9): 0, (10, 1): 0, (7, 11): 1, (0, 9): 1, (3, 7): 1, 
... (10, 3): 1, (10, 2): 1} 
>>> [k[1] for k, v in d.items() if k[0] == 10 and v == 1] 
[4, 0, 3, 2]